# $\lim\limits_{n \to \infty} \frac{a_n}{a_{n+1}} = 1 \Rightarrow \exists c \in \mathbb{R}: \lim\limits_{n \to \infty} {a_n} = c$

Let $$(a_n)_{n\in \mathbb{N}}$$ be a sequence of real numbers. I was wondering if the follwing implication is true: $$\lim\limits_{n \to \infty} \frac{a_n}{a_{n+1}} = 1 \Rightarrow \exists c \in \mathbb{R}: \lim\limits_{n \to \infty} {a_n} = c$$ Put into words: If $$\lim\limits_{n \to \infty} \frac{a_n}{a_{n+1}} = 1$$ then $$\lim\limits_{n \to \infty} a_{n}$$ converges.

My intuition behind $$\lim_{n \to \infty} \frac{a_n}{a_{n+1}} = 1$$ is that at some point $$a_n$$ and $$a_{n+1}$$ are alsmost equal. If this is the case $$(a_n)_{n\in \mathbb{N}}$$ is a Cauchy sequence and so converges. However I wasn't able to formally prove the statement. So I wondered:

• Is the statement really true?
• (If so, how could you prove it?)

Cheers,
Pascal

• $$a_n = n.{}$$ – MisterRiemann Nov 26 '18 at 22:08
• @MisterRiemann Thanks for the counter example. :) I guess I can answer the question myself then. – plauer Nov 26 '18 at 22:11

As noticed by MisterRiemann $$a_n=n$$ is a first counterexample but also $$a_n=n^2$$ works or $$a_n=\log n$$ and so on.

Therefore unfortunately your guess is definitely not true!

As a remark, other common myths on limits are:

1) $$a_n \to \infty \implies a_{n+1}\ge a_n$$

2) $$a_n \to L \implies a_n \to L^+ \quad \lor \quad a_n \to L^-$$

3) $$a_n \to 0^+ \implies a_{n+1}\le a_n$$

4) $$a_n$$ bounded $$\implies a_n \to L$$

The conclusion is wrong even for bounded sequences.

An example is $$a_ n = 2 + \sin(\log(n))$$ which is not convergent (it oscillates between $$1$$ and $$3$$). But $$\left\lvert \frac{a_{n+1}}{a_n} - 1 \right\rvert = \left\lvert \frac{\sin(\log(n+1))- \sin(\log(n))}{2 + \sin(\log(n))} \right\rvert \\ \le \left\lvert \sin(\log(n+1))- \sin(\log(n)) \right\rvert = \left\lvert \frac{\cos(\log(x_n))}{x_n}\right\rvert$$ for some $$x_n \in (n, n+1)$$, using the mean value theorem. It follows that $$\left\lvert \frac{a_{n+1}}{a_n} - 1 \right\rvert \le \frac{1}{n} \to 0 \, ,$$ i.e. $$\frac{a_{n+1}}{a_n} \to 1$$.

Roughly speaking, $$(a_n)$$ oscillates, but with decreasing frequency, so that the ratio of successive sequence elements approaches one.

Thanks to MisterRiemann for the counterexample of $$a_n = n$$.

$$\lim\limits_{n \to \infty} \frac{a_n}{a_{n+1}} = \lim\limits_{n \to \infty} \frac{n}{{n+1}} = 1$$ but: $$\lim\limits_{n \to \infty} {a_n} =\lim\limits_{n \to \infty} n = \infty \Rightarrow \nexists c \in \mathbb{R}: \lim\limits_{n \to \infty} {a_n} = c$$